So I want to find a degree $7$ polynomial whose Galois group is cyclic of order $7$. I know that to do this, I need to find a polynomial which is irreducible mod every prime (not dividing the discriminant). But this is where I'm stuck, since I don't know how to prove that any particular polynomial is irreducible mod every prime. Could I have some hints?

  • $\begingroup$ Balarka, the problem is that $\phi(n)$ is even unless $n=1$ or $2$, so $\phi(n)\neq 7$, so I don't think I can find a cyclotomic field. @John Most of the irreducibility criteria I know are for over $\mathbb Q$ or for polynomials of small degree. I also know that, for a particular prime, how to construct irreducible polynomials of a certain degree, but I'm not sure how to do it for all primes simultaneously. $\endgroup$ – Nishant Jun 16 '14 at 18:09
  • $\begingroup$ Yes, I know that, but I need irreducibility over $F_p$, not just over $\mathbb Q$. $\endgroup$ – Nishant Jun 16 '14 at 18:10
  • $\begingroup$ Hmm, since $\phi(49)=42$, would the minimal polynomial of some element of an index $6$, degree $74$ subfield of $\mathbb Q(\zeta_{49})$ work? $\endgroup$ – Nishant Jun 16 '14 at 18:25

Let $\zeta = \exp(2\pi i/29)$ be a primitive 29th root of unity. Then the field $\mathbb Q(\zeta)/\mathbb Q$ has Galois group $$ G = (\mathbb Z/29\mathbb Z)^\times \simeq \mathbb Z/28 \mathbb Z \simeq \mathbb Z/4 \mathbb Z \times \mathbb Z/7 \mathbb Z . $$ Since 2 is a primitive root mod 29 (a generator of $(\mathbb Z/29\mathbb Z)^\times$), this Galois group is generated by the automorphism $\zeta \mapsto \zeta^2$. Furthermore, an element of order 4 can by found by the seventh power of this automorphism, namely $\zeta \mapsto \zeta^{2^7} = \zeta^{12}$. Let $H$ be the subgroup of $G$ generated by $\zeta \mapsto \zeta^{12}$, which is cyclic of order 4. Then $G/H \simeq \mathbb Z/7\mathbb Z$ is the Galois group of the fixed field $K := \mathbb Q(\zeta)^H$ over $\mathbb Q$.

Now all we have to do is find the minimal polynomial of a generator for $K/\mathbb Q$. Since $\zeta$ generates $\mathbb Q(\zeta)/\mathbb Q$, the traces $\operatorname{Tr}_H(\zeta^i)$ generate $K/\mathbb Q$. Let $$ \xi = \operatorname{Tr}_H(\zeta) = \sum_{\sigma \in H} \sigma(\zeta) = \zeta + \zeta^{12} + \zeta^{28} + \zeta^{17} . $$ By solving a large system of linear equations involving the powers of $\xi$ (or using a computer to do so), we find that $\xi$ has minimal polynomial $$ f(x) = x^7 + x^6 - 12x^5 - 7x^4 + 28x^3 + 14x^2 - 9x + 1 . $$

Remark: I chose 29 because it is the smallest prime which is 1 mod 7.

Your comment above about needing the polynomial to be reducible mod every prime except those dividing the discriminant is not true. The only primes dividing the discriminant of $f$ are 17 and 29, but $f$ splits completely mod 59 or 233, for example. This is not a coincidence. These primes are both 1 mod 29, and so they split completely in $\mathbb Q(\zeta)$, and hence also in $K$

  • $\begingroup$ (+1)In general this is the inverse galois construction that can be generalized to $\Bbb Z/n\Bbb Z$. $\endgroup$ – Balarka Sen Jun 16 '14 at 18:35
  • $\begingroup$ Awesome, thanks! But why do we know that the trace generates $K$? $\endgroup$ – Nishant Jun 16 '14 at 18:38
  • $\begingroup$ @Nishant See Bruno Joyal's answer to this question. $\endgroup$ – Dane Jun 16 '14 at 18:41

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